Jae Choon Cha
POSTECH
Statements of fundamental theorems, part II
We continue assuming that $p\colon (E,e_0) \to (B,b_0)$ is a covering.
Theorem (Lifting criterion). Suppose $X$ is connected and locally path connected. For a map $f\colon (X,x_0)\to (B,b_0)$, there is a lift $\tilde f\colon (X,x_0) \to (E,e_0)$ of $f$ if and only if $f_*\pi_1(X,x_0) \subset p_* \pi_1(E,e_0)$.
The next theorem provides us a complete algebraic classification of coverings of a given base space. For this purpose, we introduce an equivalence relation on coverings.
Definition. Two coverings $p_1\colon E_1\to B$ and $p_2\colon E_2\to B$ are equivalent if there is a homeomorphism $f\colon E_1\to E_2$ such that $p_2 f = p_1$.
Definition. A space $B$ is semisimply connected if for every $x\in B$, there is an open neighborhood $U$ of $x$ such that the inclusion $i\colon U\hookrightarrow X$ induces a trivial homomorphism $i_*\colon \pi_1(U,x)\to \pi_1(X,x)$.
Theorem (Classification of coverings). Suppose $B$ is connected and locally path connected. Then the following hold:

Determination. Two coverings $p_1\colon E_1\to B$ and $p_2\colon E_2\to B$ with connected $E_1$ and $E_2$ are equivalent if and only if the subgroups $(p_1)_*\pi_1(E_1,e_1)$ and $(p_2)_*\pi_1(E_2,e_2)$ are conjugate for some $b_0\in B$ and $e_i\in p_i^{1}(b_0)$, $i=1,2$.

Realization. Suppose $B$ is, in addition, semisimply connected. Fix $b_0\in B$ arbitrarily. Then for any subgroup $H$ in $\pi_1(B,b_0)$, there is a covering $p\colon (E,e_0) \to (B,b_0)$ with $E$ connected such that $p_*\pi_1(E,e_0)=H$.
From this it follows that there is a bijection
under which the equivalence class of $p\colon E\to B$ corresponds to the conjugacy class of $p_*\pi_1(E,e_0) \subset \pi_1(B,b_0)$ where $e_0\in p^{1}(b_0)$.
By realization statement in the above theorem, if $B$ is connected, locally path connected and semisimply connected, then there exists a cover described in the following definition:
Definition. A space $X$ is simply connected if $X$ is path connected and the fundamental group of $X$ is trivial.
Definition. A cover $p\colon E\to B$ is said to be a universal cover if $E$ is simply connected.
Corollary. If $B$ is connected, locally path connected and semisimply connected, then there exists a universal cover $p\colon E\to B$.
From the classification of coverings, a universal cover of a connected and locally path connected space $B$ is unique up to equivalence if it exists.
The next result is a classification of covering transformations. Recall that, for a covering $p\colon E\to B$, a homeomorphism $f\colon E\to E$ is called a covering transformation for $p$ if $pf=p$. The set $G(EB)$ of covering transformations for $p$ is a group under composition. To state the result, we need the following notion from elementary abstract algebra: for a subgroup $H$ of a group $G$, the normalizer $N(H)$ (or $N_G(H)$) is defined by
It is straightforward to verify that (i) $N(H)$ is a subgroup of $G$ containing $H$, (ii) $H$ is a normal subgroup of $N(H)$, and (iii) $N(H)$ is the largest subgroup of $G$ satisfying (ii).
Theorem (Covering transformations and fundamental group). Suppose $E$ is connected and locally path connected, and $p\colon (E,e_0)\to (B,b_0)$ is a covering. Then there is a welldefined isomorphism of groups
where $\tilde\gamma$ designates a path in $E$ from $e_0$ to $f(e_0)$.
Proof of the theorems
Lifting criterion
Suppose there is a lift $\tilde f\colon (X,x_0)\to (E,e_0)$ of $f\colon (X,x_0)\to (B,b_0)$. Then, since $p\tilde f = f$, we have $p_* \tilde f_* = f_*$. It follows that $f_*\pi_1(X,x_0) = p_* (\tilde f_* \pi_1(X,x_0)) \subset p_*\pi_1(E,e_0)$.
To prove the converse, suppose $f_*\pi_1(X,x_0) \subset p_*\pi_1(E,e_0)$. Define a function $\tilde f\colon X\to E$ as follows. For $x\in X$, choose a path $\alpha\colon I\to X$ with $\alpha(0)=x_0$, $\alpha(1)=x$. Let $\gamma=f\alpha$, which is a path in $B$ with $\gamma(0)=b_0$. By homotopy lifting, there is a (unique) lift $\tilde\gamma\colon I \to E$ such that $\tilde\gamma(0)=e_0$. Let $\tilde f(x)=\tilde\gamma(1)$.
Welldefinedness of $\tilde f$. Suppose $\beta\colon I \to X$ is another path from $x_0$ to $x$. Let $\delta=f\beta$ and $\tilde\delta$ be the lift of $\delta$ with $\tilde\delta(0)=e_0$. We need to show that $\tilde\delta(1)=\tilde\gamma(1)$. To prove this, consider the loop $\alpha\beta^{1}$. The element $f_*[\alpha\beta^{1}] = [\gamma\delta^{1}]$ lies in $f_*\pi_1(X,x_0)$, and thus $[\gamma\delta^{1}] \in p_*\pi_1(E,e_0)$ by the hypothesis. It follows that the lift of $\gamma\delta^{1}$ starting from $e_0$ is a loop, which is of the form $\tilde\gamma * \sigma^{1}$ for some lift $\sigma$ of $\delta$ in $E$ such that $\sigma(1)=\tilde\gamma(1)$. Since $\tilde\gamma * \sigma^{1}$ is a loop based at $e_0$, we have $\sigma(0)=e_0=\tilde\delta(0)$. It follows that $\sigma =\tilde\delta$, by the uniqueness of a lift. Therefore $\tilde\delta(1)=\sigma(1)=\tilde\gamma(1)$, as desired.
Continuity of $\tilde f$. Fix $x_1\in X$. We will show that $\tilde f$ is continuous at $x_1$. Choose an evenly covered neightborhood $U$ of $f(x_1)$. Since $X$ is locally path connected, there is a path connected open neighborhood $W$ of $x_1$ in $X$ which is contained in the open set $f^{1}(U)$. Let $V$ be the sheet over $U$ which contains $f(x_1)$.
We claim that $\tilde f (x) = (p_V)^{1}\circ f$ for $x\in W$. From this claim, it follows that $f$ is continuous on $W$ since $f_W$ is a composition of continuous maps. This completes the proof. So, it only remains to prove the claim. Fix a path $\alpha_1$ in $X$ from $x_0$ to $x_1$. For each $x\in W$, choose a path $\alpha_x$ in $W$ from $x_1$ to $x$, using the path connectedness of $W$. Let $\gamma_1 = f\circ \alpha_1$, and $\tilde\gamma_1$ be the lift of $\gamma_1$ starting from $e_0$. Let $\gamma_x=f\circ \alpha_x$, and let $\tilde\gamma_x = (pV)^{1}\circ \gamma_x$. Note that $\tilde\gamma_x$ is a lift of $\gamma_x$ starting from $f(x_1)$. So, $\tilde\gamma_1 \tilde\gamma_x$ is a lift of the path $\gamma_1\gamma_x = f\circ(\alpha_1\alpha_x)$, which starts from $e_0$. Since $\alpha_1\alpha_x$ is a path in $X$ from $x_0$ to $x$, we have
This proves the claim. $\quad\square$
Remark. Observe that the hypothesis that $X$ is locally path connected plays an essential role in the proof of the continuity.
Classfication of coverings: determination
In what follows $B$ is always assumed to be locally path connected. Observe that this implies that any covering $E$ of $B$ is locally path connected too. This in turn implies that $E$ is path connected if and only if $E$ is connected.
Our goal is to show the following.
Theorem. Suppose $p_1\colon E_1\to B$ and $p_2\colon E_2\to B$ are coverings with connected $E_1$ and $E_2$. Let $b_0\in B$. Then the following are equivalent:
 The coverings $p_1$ and $p_2$ are equivalent.
 For some $e_i\in p_i^{1}(b_0)$, $i=1,2$, the subgroups $(p_1)_*\pi_1(E_1,e_1)$ and $(p_2)_*\pi_1(E_2,e_2)$ are equal.
 For some $e_i\in p_i^{1}(b_0)$, $i=1,2$, the subgroups $(p_1)_*\pi_1(E_1,e_1)$ and $(p_2)_*\pi_1(E_2,e_2)$ are conjugate.
 For all $e_i\in p_i^{1}(b_0)$, $i=1,2$, the subgroups $(p_1)_*\pi_1(E_1,e_1)$ and $(p_2)_*\pi_1(E_2,e_2)$ are conjugate.
The determination statement in the classification of coverings is an immediate consequence of this assertion.
We will first prove the following special case.
Proposition. Suppose $p_1\colon E_1\to B$ and $p_2\colon E_2\to B$ are coverings with connected $E_1$ and $E_2$. Fix $b_0\in B$, $e_1\in p_1^{1}(b_1)$ and $e_2\in p_2^{1}(b_2)$, that is, $p_i\colon (E_i, e_i) \to (B,b_0)$ for $i=1,2$. Then, there is a homeomorphism $f\colon (E_1,e_1)\to (E_2,e_2)$ such that $p_2f=p_1$ if and only if $(p_1)_* \pi_1(E_1,e_1) = (p_2)_* \pi_1(E_2,e_2)$.
Proof. Suppose there exists $f$. Then, since $p_2 f = p_1$ and $f_*\pi_1(E_1,e_1)=\pi_1(E_2,e_2)$, we have $(p_1)_* \pi_1(E_1,e_1) = (p_2)_* f_* \pi_1(E_1,e_1) = (p_2)_* \pi_1(E_2,e_2)$.
For the converse, suppose $(p_1)_* \pi_1(E_1,e_1) = (p_2)_* \pi_1(E_2,e_2)$. By the lifting criterion, there are maps $f\colon (E_1,e_1) \to (E_2,e_2)$ and $g\colon (E_2,e_2) \to (E_1,e_1)$ such that $p_2 f = p_1$ and $p_1 g = p_2$.
Observe that $p_1(gf) = (p_1g)f = p_2f = p_1 = p_1\id_{E_1}$. Since $gf(e_1)=g(e_2)=e_1=\id_{E_1}(e_1)$, it follows that $gf=\id_{E_1}$ by the uniqueness of a lift. Exchanging the roles, we obtain $fg=\id_{E_2}$. It follows that $f$ is a homeomorphism with inverse $g$. $\quad\square$
A consequence of this proposition is that (1) and (2) in the above theorem are equivalent. Indeed, if (1) holds, namely if there is a homeomorphism $f\colon E_1\to E_2$ satisfying $p_2f = p_1$, then by picking an arbitrary $e_1 \in p_1^{1}(b_0)$ and letting $e_2=f(e_1)$, (2) is satisfied by the proposition. Conversely, if (2) holds, then there is a homeomorphism $f\colon E_1\to E_2$ such that $p_2f=p_1$ by the propositon.
The above proposition leads us to investigate the effect of basepoint change for coverings. The following lemma gives an answer: briefly speaking, basepoint change for $E$ in the fiber $p^{1}(b_0)$ corresponds to conjugation on $p_*\pi_1(E,)$ in the group $\pi_1(B,b_0)$.
Lemma. Suppose $p\colon E\to B$ is a covering and $b_0\in B$ is a fixed basepoint. If $\gamma$ is a loop in $B$ based at $b_0$, and if $\tilde\gamma$ is a lift of $\gamma$ in $E$, then $p_* \tilde\gamma_* = \gamma_* p_*$, that is, the following diagram is commutative.
In particular, $[\gamma] p_*\pi_1(E,\tilde\gamma(1)) [\gamma]^{1} = p_*\pi_1(E,\tilde\gamma(0))$ in $\pi_1(B,b_0)$.
The proof of this lemma is routine and straightforward. We provide its detail below.
Proof of the lemma. Let $\tilde\alpha$ be a loop in $E$ based at $\tilde\gamma(1)$. Then we have $p_*(\tilde\gamma_*[\tilde\alpha]) = p_*[\tilde\gamma\tilde\alpha\tilde\gamma^{1}] = [\gamma (p\circ\tilde\alpha) \gamma^{1}] = \gamma_*(p_*[\tilde\alpha]).$ $\quad\square$
Proof of the theorem. We already proved that (1) and (2) are equivalent using the above proposition. We will show (2), (3) and (4) are equivalent.
It is straightforward that (2) implies (3). Suppose (3) holds for some $e_1\in p_1^{1}(b_0)$ and $e_2\in p_2^{1}(b_0)$. Let $e_1'\in p_1^{1}(b_0)$ and $e_2'\in p_2^{1}(b_0)$ be arbitrarily given. Then, by the above lemma, $(p_i)_* \pi_1(E_i,e_i')$ is conjugate to $(p_i)_ * \pi_1(E_i,e_i)$ for $i=1,2$. Since $(p_1)_ * \pi_1(E_1,e_1)$ and $(p_2)_ * \pi_1(E_2,e_2)$ are conjugate by the hypothesis, it follows that $(p_1)_ * \pi_1(E_1,e_1')$ and $(p_2)_ * \pi_1(E_2,e_2')$ are conjugate. This shows that (3) implies (4). Obviously (4) implies (3). So it suffices to show that (3) implies (2). Suppose (3) holds, that is, $(p_1)_ * \pi_1(E_1,e_1) = [\gamma] (p_2)_ * \pi_1(E_2,e_2) [\gamma]^{1}$ for some $e_i\in p_i^{1}(b_0)$ and some $[\gamma]\in \pi_1(B,b_0)$. Let $\tilde\gamma$ be the lift of $\gamma$ satisfying $\tilde\gamma(1)=e_2$. By the lemma,
Therefore, for $e_2' = \tilde\gamma(0)$, we have $(p_1)_ * \pi_1(E_1,e_1) = (p_2)_* \pi_1(E_2,\tilde\gamma(0))$. This shows that (3) implies (2).
Classification of coverings: realization
Recall that our goal is to prove the following statement: if $B$ is connected, locally path connected and semilocally simply connected, and if $H$ is a subgroup of $\pi_1(B,b_0)$, then there is a covering $p\colon (E,e_0)\to (B,b_0)$ with $E$ connected which satisfies $p_*\pi_1(E,e_0)=H$.
Here, the idea is as follows. Just hypothetically, suppose we had a desired covering $p\colon (E,e_0) \to (B,b_0)$. Then, we already know that the fiber $p^{1}(b_0)$ is in 11 correspondence with the right cosets of $H$. That is,

A loop $\alpha$ in $B$ based at $b_0$ gives a point in the fiber, that is, the point $\tilde\alpha(1)\in p^{1}(b_0)$ is associated to $\alpha$, where $\tilde\alpha$ is the lift starting from $e_0$.

Every point in the fiber $p^{1}(b_0)$ can be given in this way.

Two loops $\alpha$, $\beta$ give the same point if and only if $[\alpha\beta^{1}]$ lies in $H$.
Recall that associating a point to a loop (namely, $\alpha\mapsto \tilde\alpha(0)$) is useful, for instance for the purpose of understanding $\pi_1$ in terms of points in the fiber, as exhibited in the computation of $\pi_1(S^1)$. On the other hand, the opposite direction, namely associating a loop to a point, gives us the the possibility of understanding points in terms of loops. That is, when you don’t know what the fiber $p^{1}(b_0)$ is, how can you recover $p^{1}(b_0)$ from the fundamental group information only? Namely from the subgroup $H=p_*\pi_1(E,e_0)$ of $\pi_1(B,b_0)$? The answer is that you can just take the right cosets of $H$!
Now, let’s try to generalize this, to not only points in $p^{1}(e_0)$ but also all points in $E$. Recall that the proof of the second statement above proceeds as follows: for a point $e\in p^{1}(b_0)$, choose a path $\gamma$ from $e_0$ to $e$, and let $\alpha=p\circ\gamma$, the projection of $\gamma$. This $\alpha$ is the loop which gives the point $e$. To generalize this, we will to apply the same idea when $e$ is an arbitrary point in $E$. Choose a path $\gamma$ in $E$, from $e_0$ to the given $e$, and let $\alpha=p\circ\gamma$. An important difference is that now $\alpha$ is no longer a loop! It is just a path starting from $b_0$. Nevertheless, we still want to proceed similarly to the case of points in $p^{1}(b_0)$. When a path $\alpha$ in $B$ starting from $b_0$ is given, take a lift $\tilde\alpha$ satisfying $\tilde\alpha(0)=e_0$, and associate the point $\tilde\alpha(0)$ to $\alpha$. In this way, we want to think of analogs of the above #1 and #2, to relate points $e\in E$ and paths $\alpha$ in $B$ starting from $e_0$.
So, it is natural to ask whether we can also generalize #3 above. That is, exactly when two given paths (not loops!) $\alpha$ and $\beta$ in $B$ with $\alpha(0)=b_0=\beta(0)$ gives us the same point in $E$? Recall that $\alpha$, $\beta$ give the points $\tilde\alpha(1)$, $\tilde\beta(1) \in E$. So, when do we have $\tilde\alpha(1)=\tilde\beta(1)$? If it is the case, then $\tilde\alpha \tilde\beta^{1}$ is defined, and $\tilde\alpha \tilde\beta^{1}$ is a loop based at $e_0$. In addition, $\alpha\beta^{1}$ is defined too, and $\alpha\beta^{1}$ is a loop based at $b_0$. Moreover, the element $[\alpha\beta^{1}]$ in $\pi_1(B,b_0)$ lies in $H=p_*\pi_1(E,e_0)$, since $\alpha\beta^{1}= p\circ(\tilde\alpha \tilde\beta^{1})$. Conversely, if $\alpha$, $\beta$ are path in $B$ which start from $b_0$ and satisfy $\alpha(1)=\beta(1)$ and $[\alpha\beta^{1}]\in H$, then it is not difficult to see that $\tilde\alpha(1)=\tilde\beta(1)$. This says that the points in $E$ associated to $\alpha$ and $\beta$ are the same if and only if $\alpha(1)=\beta(1)$ and $[\alpha\beta^{1}]\in H$.
From the above discussion, we see that $E$, as a set (not as a space for now), is in 11 correspondence with the set of equivalence classes
where $\alpha\sim \beta$ if and only if $[\alpha\beta^{1}]\in H$.
Recall that this is done when a covering $p\colon (E,e_0)\to (B,b_0)$ is given. But, in our original situation, we do not have $E$. We are given the base space $B$ and a subgroup $H \subset \pi_1(B,b_0)$ only, and want to construct a covering $E$ whose $\pi_1$ is $H$. The above discussion motivates how we proceed to construct $E$ from $H$: take the above set of equivalence classes, which is completely given from $B$ and $H$, as your $E$. All right, but isn’t $E$ just a set, not a space? In the following rigourous proof, you will see how to construct an appropriate topology on $E$ as well. The main idea of our construction of the topology is that if we restrict ourselves to an evenly covered neighborhood and a sheet on it, then they must have the same open sets. So, briefly speaking, the topology of $E$ must be specified from what the open sets of $B$ are. Let’s proceed to a rigourous proof!
Proof of the second part of the classification of coverings. Suppose $B$ is connected, locally path connected and semilocally simply connected, and suppose $H$ is a subgroup of $\pi_1(B,b_0)$. We will construct a covering $p\colon (E,e_0)\to (B,b_0)$ with $E$ connected which satisfies $p_*\pi_1(E,e_0)=H$.
Consider the set of paths in $B$ starting from $b_0$. On this set, define a relation by $\alpha\sim\beta$ if $\alpha(1)=\beta(1)$ and $[\alpha\beta^{1}] \in H$. It is straightforward to verify that $\sim$ is an equivalence relation. Let $\mathcal P$ be the set of equivalence classes. That is,
Denote the equivalence class of $\alpha$ by $\langle\alpha\rangle$. (This should be distinguished from the notation $[\alpha]$, which designates the homotopy class rel $\{0,1\}$. Especially $[\alpha]$ is meaningful only when $\alpha$ is a loop, while $\langle\alpha\rangle$ makes sense when $\alpha$ is a path starting from $b_0$.)
Note that $\langle\alpha\rangle= \langle\beta\rangle$ if $\alpha\simeq\beta \rel \{0,1\}$.
Define a function $p\colon \mathcal P \to B$ by $p(\langle\alpha\rangle)=\alpha(1)$. This is obviously welldefined.
Now we define a topology on $\mathcal P$ as follows. First we introduce a new notation. For $\alpha\colon (I,0) \to (B,b_0)$ and an open neighborhood $V$ of $\alpha(1)$, let
Claim 1. The sets $\langle \alpha,V \rangle$ form a base for a topology on the set $\mathcal P$.
Proof of Claim 1: Recall that a set $\mathcal B$ of subsets of a set $X$ is a base of a topology on $X$ if and only if the following hold: (1) $\bigcup_{V\in \mathcal B}V=X$, and (2) whenever $x\in V\cap W$ and $V,W\in \mathcal B$, there is $U\in B$ such that $x \in U \subset V\cap W$. In our case, for each $\langle \alpha\rangle \in \mathcal P$, $\langle\alpha\rangle \in \langle \alpha,V\rangle$ for every open neighborhood $V$ of $\alpha(1)$. So $\bigcup \langle\alpha,V\rangle = \mathcal P$. Also, if $\langle\gamma\rangle \in \langle\alpha,V\rangle \cap \langle\beta,W\rangle$, then we have $\langle\gamma\rangle \in \langle\gamma,V\cap W\rangle \subset \langle\alpha,V\rangle \cap \langle\beta,W\rangle$. This shows the claim.
From now on, regard $\mathcal P$ as a topological space equipped with the topology with the base $\{\langle \alpha,V \rangle\}$.
Claim 2. The function $p\colon \mathcal P \to B$ is continuous and open.
Proof of Claim 2: For an open set $V\subset B$, $p^{1}(V)=\{\langle\alpha\rangle\mid \alpha(1)\in V\}$. So, if $\langle\alpha\rangle \in p^{1}(V)$, then $\langle\alpha\rangle \in \langle\alpha,V\rangle \subset p^{1}(V)$. This shows that $p$ is continuous. Moreover, $p(\langle\alpha,V\rangle)$ is equal to the path component of $V$ containing $\alpha(1)$, and this is open since $B$ is locally path connected. It follows that $p$ is an open map.
Claim 3. If $V\subset B$ is open and path connected and if $\pi_1(V,*)\to \pi_1(B,*)$ is trivial for some $*\in V$, then $V$ is evenly covered.
Before the proof, observe that the hypothesis of Claim 3 imples that $\pi_1(V,*)\to \pi_1(B,*)$ is trivial for all $*\in V$, since $V$ is path connected.
Proof of Claim 3: If $p(\langle\alpha\rangle) = \alpha(1)\in V$, then $p(\langle \alpha,V\rangle) = V$. So we have $p^{1}(V) = \bigcup_{\alpha(1)\in V} \langle \alpha,V\rangle$. (In this union, $\alpha$ varies over all $\alpha\colon (I,0)\to (B,b_0)$ such that $\alpha(1)\in V$.) Also, if $\langle\alpha,V\rangle \cap \langle\beta,V\rangle$ is nonempty, then $\langle\alpha,V\rangle = \langle\beta,V\rangle$ since $V$ is path connected. It follows that the above union is a union of disjoint open sets. To show that each $\langle \alpha,V\rangle$ is a sheet whenever $\alpha(1)\in V$, we need to verify that $p_ {\langle\alpha,V\rangle}$ is a homeomorphism onto $V$. We already verified that $p_ {\langle\alpha,V\rangle}$ is onto. To show that $p_ {\langle\alpha,V\rangle}$ is 11, suppose $p(\langle\alpha\beta\rangle) = p(\langle\alpha\gamma\rangle)$ for some $\beta,\gamma\colon (I,0) \to (V,\alpha(1))$. Then $\beta(1)=\gamma(1)$, and $[\beta\gamma^{1}] \in \pi_1(V,\alpha(1))$. By the hypothesis, $[\beta\gamma^{1}]$ is trivial in $\pi_1(B,\alpha(1))$, that is, $\beta\gamma^{1} \simeq c_{\alpha(1)} \rel\{0,1\}$ in $B$. It follows that $\beta \simeq \beta\gamma^{1}\gamma\simeq \gamma \rel\{0,1\}$ in $B$, and consequently $\langle\alpha\beta\rangle=\langle\alpha\gamma\rangle$. This shows that $p_ {\langle\alpha,V\rangle}$ is 11. Since $p$ is continuous and open by Claim 2, from this it follows that $p_ {\langle\alpha,V\rangle}$ is a homeomorphism. This completes the proof of Claim 3.
From Claim 3, it follows that $p\colon \mathcal P \to B$ is a covering map, since $B$ is assumed to be semilocally simply connected.
Claim 4. $\mathcal P$ is path connected.
Proof of Claim 4: Let $e_0 = \langle c_{b_0} \rangle \in \mathcal P$. For $\langle\alpha\rangle\in \mathcal P$ and $s\in I$, define $\alpha_s\colon I \to B$ by $\alpha_s(t)=\alpha(st)$. Define $\tilde\alpha\colon I\to \mathcal P$ by $\tilde\alpha(s)=\langle \alpha_s \rangle$. Then $(p\circ \tilde\alpha)(s) = p(\langle\alpha_s\rangle) = \alpha_s(1)=\alpha(s)$. This shows that $p\tilde\alpha=\alpha$, that is, $\tilde\alpha$ is a lift of $\alpha$.
To show that $\tilde\alpha$ is continuous, fix $s_0\in I$, choose a neighborhood $V$ of $\alpha(s_0)$ in $B$ which satisfies the hypothesis of Claim 3, and choose a subinterval $J=(s_0\epsilon,s_0+\epsilon)\cap I$ ($\epsilon>0$) such that $\alpha(J)\subset V$. Then, for $s\in J$, $\tilde\alpha(s)$ lies in the sheet $\langle\alpha_{s_0},V\rangle$. So, on $J$, $\tilde\alpha=(p_ {\langle\alpha_{s_0},V\rangle})^{1}\circ \alpha$ is continuous.
Observe that $\tilde\alpha(0)=\langle \alpha_0 \rangle = \langle c_{b_0} \rangle = e_0$ and $\tilde\alpha(1)= \langle \alpha_1 \rangle = \langle \alpha \rangle$. So $\tilde\alpha$ is a path from $e_0$ to $\langle \alpha \rangle$. This completes the proof of Claim 4.
Claim 5. $p_*\pi_1(\mathcal P,e_0) = H \subset \pi_1(B,b_0)$.
Proof of Claim 5: Let $[\alpha]\in \pi_1(B,b_0)$. Let $\tilde\alpha$ be the lift of $\alpha$ which starts from $e_0$. By the uniqueness of a lift, this $\tilde\alpha$ is the one given in the proof of Claim 3. In particular, $\tilde\alpha(1)=\langle\alpha\rangle$. Therefore,
This completes the proof of Claim 5.
By the above claims, $p\colon (\mathcal P,e_0) \to (B,b_0)$ is a connected covering such that $p_*\pi_1(\mathcal P,e_0) = H$. $\quad\square$
Covering transformations and fundamental group
Suppose $E$ is connected and locally path connected, and $p\colon (E,e_0)\to (B,b_0)$ is a covering. The goal is to prove that
is a welldefined isomorphism between groups. Here, for a covering transformation $f$ on the left side, $\tilde\gamma$ on the right side is a path in $E$ from $e_0$ to $f(e_0)$.
Proof. First, we verify that $[p\circ\tilde\gamma]$ lies in the normalizer $N(p_*\pi_1(E,e_0))$. Let $\gamma=p\circ\tilde\gamma$. We have
where the last equality follows from $pf = p$. This shows that $[p\circ\tilde\gamma] \in N(p_*\pi_1(E,e_0))$.
Next, we claim that $\Phi$ is well defined, indepdent of the choice of $\tilde\gamma$. Suppose $\tilde\delta$ is another path in $E$ from $e_0$ to $f(e_0)$, and let $\delta=p\circ\tilde\delta$. Then $[\gamma][\delta]^{1} = p_*[\tilde\gamma\tilde\delta^{1}]\in p_*\pi_1(E,e_0)$. So $[\gamma]$ and $[\delta]$ are in the same coset, as we claimed.
To verify that $\Phi$ is a homomorphism, consider two covering transformations $f$ and $g$ in $G(EB)$. Let $\tilde\gamma$ and $\tilde\delta$ be paths in $E$ from $e_0$ to $f(e_0)$ and $g(e_0)$ respectively. Then the product $\tilde\gamma * (f\circ \tilde\delta)$ is defined, and it is a path from $e_0$ to $(f\circ g)(e_0)$. So, $\Phi(f\circ g)$ is the coset of $[p\circ (\tilde\gamma * (f\circ \tilde\delta))]$, which is equal to $[p\tilde\gamma]\cdot[p\tilde\delta]$ since $pf=p$. It follows that $\Phi(f\circ g)$ = $\Phi(f)\Phi(g)$.
Now it remains to show that $\Phi$ is 11 and onto. Suppose $f\in G(EB)$, and take $\tilde\gamma$ which is from $e_0$ to $f(e_0)$. If $\Phi(f)$ is trivial, then $[p\circ\tilde\gamma] \in p_*\pi_1(E,e_0)$, and thus $\tilde\gamma$ is a loop (by the uniqueness of a lift). It follows that $f(e_0)=\tilde\gamma(1)=\tilde\gamma(0)=e_0$. By the uniqueness of a lift (viewing $f$ as a lift of $p$), $f$ is the identity. This shows that $\Phi$ is 11. Finally, to show that $\Phi$ is onto, let $[\gamma]$ be an element in the normallizer $N(p_*\pi_1(E,e_0))$. Let $\tilde\gamma$ be the lift of $\gamma$ with $\tilde\gamma(0)=e_0$. Then we have
Using this, apply the lifting criterion to the following situation
to obtain a lift $f\colon (E,e_0) \to (E,\tilde\gamma(1))$, and an opposite direction lift $g\colon (E,\tilde\gamma(1)) \to (E,e_0)$. By the uniqueness of a lift, it follows that $gf$ and $fg$ are the identity on $E$. Therefore $f$ is a covering transformation satisfying $f(e_0)=\tilde\gamma(1)$, that is, $\Phi(f)$ is equal to the coset of $[\gamma]$. This shows that $\Phi$ is onto. $\quad\square$